It has nothing to do with prime moduli or CRT. Rather, it is true for all moduli $\,n\,$ as we can easily prove as follows, using notation $\ \bar x := x\bmod n\, $ for the remainder left after dividing $\,x\,$ by $\,n.$
Applying the equivalence $\ x\equiv y\pmod{\! n}\!\color{#c00}\iff \bar x = \bar y,\, $ and $\,\small \rm\color{#0a0}{CPR} =$ Congruence Product Rule
$$\begin{align} {\rm mod}\ n\!:\,\ a \,&\color{#c00}\equiv\, \bar a\\
b\,&\color{#c00}\equiv\, \bar b\\
\smash{\overset{\rm\color{#0a0}{CPR}}\Longrightarrow}\,\ \ a\:\!b
\,&\equiv\, \bar a\:\!\bar b\\
\color{#c00}\Longrightarrow\ a\:\!b\bmod n \,&=\, \bar a\:\! \bar b\bmod n,\ \ {\rm i.e.}\ \ \overline{ab} = \overline{\bar a \bar b} \end{align}\!\!$$
Remark $ $ Generally, as above, to prove an identity about mod as an operator it is usually easiest to first convert it into the more flexible congruence form, using $\color{#c00}\Longleftarrow$ in the equivalence, then prove it using congruences, then at the end, convert it back to operator form, using $\color{#c00}\Longrightarrow$ in the equivalence.
For a similar example consider equivalence of fractions, $\,\frac{a}b \equiv \frac{c}d\iff ad = bc.\,$ It would be quite cumbersome to do fraction arithmetic if we required that all fractions always be in normal (reduced) form, i.e. in least terms. Instead, it proves more convenient to have the flexibility to work with arbitrary equivalent fractions. For example, this allows us to state the fraction addition rule in very simple form by first choosing convenient reps having a common denominator.
See here for further discussion on $\!\bmod\!$ as a binary operator vs. equivalence relation.
See here for how to interpret the above as transporting the ring operations on cosets in the quotient ring $\,\Bbb Z/n\,$ to corresponding operations on their normal-form (remainder) reps in $\,\Bbb Z_n$.
Best Answer
Since everything is $\bmod{26}$, you can use most of the methods for solving other simultaneous equations. Instead of dividing to get fractions, use modular division (which involves the Euclideam Algorithm).
For example, let's use Gaussian elimination for this problem $$ \begin{align} 12&=2a+b\pmod{26}\\ 15&=9a+b\pmod{26} \end{align} $$ Subtracting the first from the second gives $$ 3=7a\pmod{26} $$ Using the Euclidean Algorithm, we get that $15\times7=105\equiv1\pmod{26}$. So, multiplying both sides by $15$ we get $$ 19=a\pmod{26} $$ Subtracting $2$ times the second from $9$ times the first yields $$ 78=7b\pmod{26} $$ Since $78\equiv0\pmod{26}$, multiplying both sides by $15$ yields $$ 0=b\pmod{26} $$
Using the Euclid-Wallis Algorithm
As described in this answer, we can use the Euclid-Wallis Algorithm to invert $7\bmod{26}$: $$ \begin{array}{rrrrrrr} &&\color{orange}{3}&\color{orange}{1}&\color{orange}{2}&\color{orange}{2}\\ \hline \color{#00A000}{1}&\color{#00A000}{0}& 1&-1&\color{red}{3}&\color{blue}{-7}\\ \color{#00A000}{0}& \color{#00A000}{1}& -3&4&\color{red}{-11}&\color{blue}{26}\\ \color{#00A000}{26}&\color{#00A000}{7}&5& 2& \color{red}{1}&\color{blue}{0} \end{array} $$ This says that $3\times26-11\times7=1$, which says that $-11\times7\equiv1\pmod{26}$. Since $-11\equiv15\pmod{26}$, we also get $15\times7\equiv1\pmod{26}$.